如何用一个自动生成的值创建复合主键？

busg9geu 于 2021-06-20 发布在 Mysql

关注(0)|答案(2)|浏览(517)

I am trying to map entity with my existing database table which is having two primary keys.

Out of two keys, one primary key is auto generated.

I am using `Spring Boot`, `JPA`, `Hibernate` and `MySQL`. I have used `@IdClass` to map with the composite primary class (with public constructor, setter/getters, same property names, equals and hash code methods).

`org.springframework.data.repository.CrudRepository` save method to save the entity.

下面的代码段。

@Entity
@Table(name = "data")
@IdClass(DataKey.class)
public class DeviceData {
    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name="id")
    private BigInteger id;
    @Column(name="name")
    private String name;
    @Id
    @Column(name="device_id")
    private int deviceId;
    getters/setters
}
public class DataKey implements Serializable {
    private static final long serialVersionUID = 1L;
    private BigInteger id;
    private int deviceId;
    //setter/getters
    public DataKey() {
    }
    public DataKey(BigInteger id, int deviceId) {
        this.id = id;
        this.deviceId = deviceId;
    }
    public int hashCode() {
        return Objects.hash(id, deviceId);
    }
    public boolean equals(Object obj) {
        if (obj == this)
            return true;
        if (!(obj instanceof DataKey))
            return false;
        DataKey dk = (DataKey) obj;
        return dk.id.equals(this.id) && dk.deviceId == (this.deviceId);}
}
I am using org.springframework.data.repository.CrudRepository save method for persisting the entity.

devicedata data=new devicedata（）；
data.setname（“设备1”）；data.setdeviceid（“1123”）；
保存（数据）//道扩展了粗糙的存储接口。
但我得到以下错误：

org.springframework.orm.jpa.JpaSystemException: Could not set field 
    value [POST_INSERT_INDICATOR] value by reflection.[class DataKey.id]
    setter of DataKey.id; nested exception is
    org.hibernate.PropertyAccessException: Could not set field value
    [POST_INSERT_INDICATOR] value by reflection : [class class DataKey.id]
    setter of class DataKey.id.

原因：java.lang.illegalargumentexception:无法将java.math.biginteger字段datakey.id设置为org.hibernate.id.identifiergeneratorhelper$2

mysql Hibernate jpa composite-key autogeneratecolumn

来源：https://stackoverflow.com/questions/51392223/how-to-create-composite-primary-key-with-one-auto-generated-value

2条答案

按热度按时间

camsedfj1#

我在我的一个项目中也尝试过同样的方法，但最终我发现在jpa中自动生成一个键属性（在复合主键的情况下）是不可能的。所以，我分别通过调用序列和查询来完成。

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nzkunb0c2#

I have achieved this by creating a one orderId class with parameters -- Mysql
@Getter
@Setter
public class OrderId implements Serializable {

    private int oId;
    private int customerId;
    private int productId;

    @Override
    public boolean equals(Object o) {
        if (this == o)
            return true;
        if (o == null || getClass() != o.getClass())
            return false;
        OrderId orderId = (OrderId) o;
        return oId == orderId.oId && customerId == orderId.customerId && productId == orderId.productId;
    }

    @Override
    public int hashCode() {
        return Objects.hash(oId, customerId, productId);

    }

the class OrderData with orderid as auto generated field in mysql
@Entity
@Table(name = "OrderData")
@Getter
@Setter
@NoArgsConstructor
@IdClass(OrderId.class)
public class OrderData{
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int oId;
    @Id
    private int customerId;
    @Id
    private int productId;
    private int quanity;
    private String paymentMode;
    private String orderStatus;

}

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如何用一个自动生成的值创建复合主键？

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